Re: Abstract algebra, short problems

your answer for number 1 is correct. note that the first subset is the elements of <3> that are generators of <3>, <3> has order 10, so should have φ(10) = 4 generators. similarly <x4> has order 10, and will have 4 generators as well.

there is a problem with your "proof" ϕ is injective. you cannot conclude that 2x = 2y implies x = y. how do you get from step 1, to step 2? it looks like you divided by 2, but you cannot do that in Z25, unless 2 is a unit (not all elements of Z25 ARE units, for example 5 is NOT).

just for emphasis: Aut(Zn) ≅ U(n).

in this case, 2 IS a unit in Z25: (2)(13) = 26 = 1 (mod 25).

therefore, you should have proceeded like this:

2x = 2y
(13)(2x) = (13)(2y)
((13)(2))x = ((13)(2))y
x = y

there is also a problem with your "proof" that ϕ is surjective, when you say "let y = 2x". that is precisely what you want to PROVE, is that there is some x with y = 2x. so you can't ASSUME there is right from the start. for all we know the set of 2x's might be SMALLER than the set of y's. but...we already know that (2)(13) = 1 (mod 25). so choose x = 13y. this is clearly "some" element of Z25, and ϕ(x) = ϕ(13y) = 2(13y) = ((2)(13))y = 1y = y, as desired.